Question

0.055 grams of PbSO4 is dissolved in 200.00 grams of H2O.

Answer 1

Answer:

Solution's mass = 200.055 g

[PbSO₄] = 275 ppm

Explanation:

Solute mass = 0.055 g of lead(II) sulfate

Solvent mass = 200 g of water

Solution mass = Solvent mass + Solution mass

0.055 g + 200 g = 200.055 g

ppm = μg of solute / g of solution

We convert the mass of solute from g to μg

0.055 g . 1×10⁶ μg/ 1g = 5.5×10⁴μg

5.5×10⁴μg / 200.055 g = 275 ppm

ppm can also be determined as mg of solute / kg of solution

It is important that the relation is 1×10⁻⁶

Let's verify: 0.055 g = 55 mg

200.055 g = 0.200055 kg

55 mg / 0.200055 kg = 275 ppm